文章目录：
- 1、用dev c++怎么用c语言编写一个连连看游戏？写这个游戏难吗？一个星期够吗？
- 2、求连连看源代码
- 3、求任何可在VC/C++环境下运行的连连看游戏完整源代码文件 谢谢了。。
- 4、求c语言的连连看源程序
- 5、谁能帮忙写一个c语言连连看游戏，要4x4的，功能最简单的就可以！但是要实现连连看应该有的功能
- 6、急需一个用C语言编写的 连连看游戏 代码啊，需要简易一点的，适合刚入门新手看懂的
用dev c++怎么用c语言编写一个连连看游戏？写这个游戏难吗？一个星期够吗？
连的元素是什么？ 图片？还是纯字母？
难度不高，不知你水平如何，如果是我写的话，应该几个小时就行了。
求连连看源代码
连连看的代码（基本算法）加了部分注释
import java.awt.*;
import java.awt.event.*;
public class lianliankan implements ActionListener
{
static String s="no"; //用来纪录点击按钮的信息
int x0=0,y0=0,x=0,y=0,n1=0,n2=0; //用来纪录按钮的位置信息
Frame f,f1;
Button b1,b2,b3,b4,b5,b6,b7,b8,b9,b10; //用比较笨的方法添加了
Button b11,b12,b13,b14,b15,b16,b17,b18; //30个按钮来实现游戏界面
Button b19,b20,b21,b22,b23,b24,b25; //可以用数组实现，这是本人
Button b26,b27,b28,b29,b30,bc; //学java时，入门的联系，所以
Button b,ba,br,bt1,bt2; //有些东西很业余！！嘻嘻
Panel p1,p2,p3;
TextField t; //用来显示一些随机信息，方法是下面的guli().
Label l;
int d[][]={ //用来和界面的按钮建立映射关系
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0}
};
public static void main(String[] args)
{
lianliankan t=new lianliankan();
t.suiji();
t.go();
}
public void actionPerformed(ActionEvent e) //再来一次按钮的响应事件。
{
int d[][]={
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0}
};
this.d=d;
suiji();
f.setVisible(false);
f1.setVisible(false);
s="no";
go();
}
public void go()//初始化界面
{
l=new Label("亲爱的玩家，");
f=new Frame("连连看");
t=new TextField();
p2=new Panel();
p1=new Panel();
p3=new Panel();
bc=new Button("退出");
br=new Button("重列");
b=new Button();
b1=new Button(String.valueOf(d[1][1]));
b2=new Button(String.valueOf(d[1][2]));
b3=new Button(String.valueOf(d[1][3]));
b4=new Button(String.valueOf(d[1][4]));
b5=new Button(String.valueOf(d[1][5]));
b6=new Button(String.valueOf(d[2][1]));
b7=new Button(String.valueOf(d[2][2]));
b8=new Button(String.valueOf(d[2][3]));
b9=new Button(String.valueOf(d[2][4]));
b10=new Button(String.valueOf(d[2][5]));
b11=new Button(String.valueOf(d[3][1]));
b12=new Button(String.valueOf(d[3][2]));
b13=new Button(String.valueOf(d[3][3]));
b14=new Button(String.valueOf(d[3][4]));
b15=new Button(String.valueOf(d[3][5]));
b16=new Button(String.valueOf(d[4][1]));
b17=new Button(String.valueOf(d[4][2]));
b18=new Button(String.valueOf(d[4][3]));
b19=new Button(String.valueOf(d[4][4]));
b20=new Button(String.valueOf(d[4][5]));
b21=new Button(String.valueOf(d[5][1]));
b22=new Button(String.valueOf(d[5][2]));
b23=new Button(String.valueOf(d[5][3]));
b24=new Button(String.valueOf(d[5][4]));
b25=new Button(String.valueOf(d[5][5]));
b26=new Button(String.valueOf(d[6][1]));
b27=new Button(String.valueOf(d[6][2]));
b28=new Button(String.valueOf(d[6][3]));
b29=new Button(String.valueOf(d[6][4]));
b30=new Button(String.valueOf(d[6][5]));
p3.setLayout(null);
p1.setSize(250,300);
p2.setSize(100,40);
p3.setSize(300,30);
t.setSize(60,30);
l.setSize(70,30);
p1.setLayout(new GridLayout(6,5));
p1.setBackground(Color.pink);
p1.setLocation(100,100);
p2.setLocation(0,400);
p3.setLocation(50,50);
t.setLocation(230,2);
l.setLocation(150,2);
bc.setLocation(0,40);
br.setLocation(0,100);
f.add(p1);
f.add(p2);
f.add(p3);
p3.add(l);
p3.add(t);
p2.add(bc);
p2.add(br);
p1.add(b1);
p1.add(b2);
p1.add(b3);
p1.add(b4);
p1.add(b5);
p1.add(b6);
p1.add(b7);
p1.add(b8);
p1.add(b9);
p1.add(b10);
p1.add(b11);
p1.add(b12);
p1.add(b13);
p1.add(b14);
p1.add(b15);
p1.add(b16);
p1.add(b17);
p1.add(b18);
p1.add(b19);
p1.add(b20);
p1.add(b21);
p1.add(b22);
p1.add(b23);
p1.add(b24);
p1.add(b25);
p1.add(b26);
p1.add(b27);
p1.add(b28);
p1.add(b29);
p1.add(b30);
f.pack();
f.setBounds(280,100,500,450);
f.setResizable(false);
f.setVisible(true);
bc.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
ex();
}
});
br.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
chonglie();
}
});
b1.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(1,1,b1);
}
});
b2.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(1,2,b2);
}
});
b3.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(1,3,b3);
}
});
b4.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(1,4,b4);
}
});
b5.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(1,5,b5);
}
});
b6.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(2,1,b6);
}
});
b7.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(2,2,b7);
}
});
b8.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(2,3,b8);
}
});
b9.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(2,4,b9);
}
});
b10.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(2,5,b10);
}
});
b11.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(3,1,b11);
}
});
b12.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(3,2,b12);
}
});
b13.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(3,3,b13);
}
});
b14.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(3,4,b14);
}
});
b15.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(3,5,b15);
}
});
b16.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(4,1,b16);
}
});
b17.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(4,2,b17);
}
});
b18.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(4,3,b18);
}
});
b19.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(4,4,b19);
}
});
b20.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(4,5,b20);
}
});
b21.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(5,1,b21);
}
});
b22.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(5,2,b22);
}
});
b23.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(5,3,b23);
}
});
b24.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(5,4,b24);
}
});
b25.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(5,5,b25);
}
});
b26.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(6,1,b26);
}
});
b27.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(6,2,b27);
}
});
b28.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(6,3,b28);
}
});
b29.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(6,4,b29);
}
});
b30.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
wei(6,5,b30);
}
});
}
public void ex() //退出界面，可用diolog来实现有模式的类型，更加符合
{
f1=new Frame("游戏作业");
f1.setLayout(new GridLayout(1,1));
bt1=new Button("确定退出");
bt2=new Button("再来一局");
f1.add(bt1);
f1.add(bt2);
f1.pack();
f1.setBounds(400,250,90,60);
f1.setResizable(false);
f1.show();
f1.setVisible(true);
bt1.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e)
{
System.exit(0);
}
});
bt2.addActionListener(this);
}
public void suiji() //产生随机数，来填充游戏界面对应的数组的各个位置
{
int m,n,k=0,k1,k2,k3;
for(m=1;m=15;m++)
{
k1=(int)(Math.random()*25+1);
for(n=1;n=2;n++)
{
k2=(int)(Math.random()*6+1);
k3=(int)(Math.random()*5+1);
while(d[k2][k3]!=0 k!=30)
{
k2=(int)(Math.random()*6+1);
k3=(int)(Math.random()*5+1);
}
this.d[k2][k3]=k1;
k++;
}
}
}
public void guli() //随机信息
{
int l=0;
t.setText("");
l=(int)(Math.random()*10);
System.out.println(l);
switch(l)
{
case 1:
t.setText("好！加油！");
break;
case 3:
t.setText("你真棒！");
break;
case 5:
t.setText("加快速度！");
break;
case 6:
t.setText("不错啊！");
break;
case 8:
t.setText("加油吧！");
break;
case 9:
t.setText("够聪明!");
break;
default:
break;
}
}
public void chonglie() //重列方法
{
int save[],i,j,n=0,k2,k3,k;
int d[][]={
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0}
};
save=new int[30];
for(n=0;n30;n++)
save[n]=0; //定义一个数组来保存当前的每个按钮位置上的信息
n=0;
for(i=0;i=6;i++)
for(j=0;j=5;j++)
{
if(this.d[i][j]!=0)
{
save[n]=this.d[i][j];
n++;
}
}
n=n-1;
this.d=d;
while(n=0) //产生随机位置，放置按钮
{
k2=(int)(Math.random()*6+1);
k3=(int)(Math.random()*5+1);
while(d[k2][k3]!=0)
{
k2=(int)(Math.random()*6+1);
k3=(int)(Math.random()*5+1);
}
this.d[k2][k3]=save[n];
n--;
}
f.setVisible(false);
s="no"; //这里一定要将按钮点击信息归为初始
go();
ling();
}
public void ling() //将数组中为零的成员对应的按钮消去
{ //用按钮类型的数组实现会简化得多，
if(d[1][1]==0)
b1.setVisible(false);
if(d[1][2]==0)
b2.setVisible(false);
if(d[1][3]==0)
b3.setVisible(false);
if(d[1][4]==0)
b4.setVisible(false);
if(d[1][5]==0)
b5.setVisible(false);
if(d[2][1]==0)
b6.setVisible(false);
if(d[2][2]==0)
b7.setVisible(false);
if(d[2][3]==0)
b8.setVisible(false);
if(d[2][4]==0)
b9.setVisible(false);
if(d[2][5]==0)
b10.setVisible(false);
if(d[3][1]==0)
b11.setVisible(false);
if(d[3][2]==0)
b12.setVisible(false);
if(d[3][3]==0)
b13.setVisible(false);
if(d[3][4]==0)
b14.setVisible(false);
if(d[3][5]==0)
b15.setVisible(false);
if(d[4][1]==0)
b16.setVisible(false);
if(d[4][2]==0)
b17.setVisible(false);
if(d[4][3]==0)
b18.setVisible(false);
if(d[4][4]==0)
b19.setVisible(false);
if(d[4][5]==0)
b20.setVisible(false);
if(d[5][1]==0)
b21.setVisible(false);
if(d[5][2]==0)
b22.setVisible(false);
if(d[5][3]==0)
b23.setVisible(false);
if(d[5][4]==0)
b24.setVisible(false);
if(d[5][5]==0)
b25.setVisible(false);
if(d[6][1]==0)
b26.setVisible(false);
if(d[6][2]==0)
b27.setVisible(false);
if(d[6][3]==0)
b28.setVisible(false);
if(d[6][4]==0)
b29.setVisible(false);
if(d[6][5]==0)
b30.setVisible(false);
}
public void wei(int w1,int w2,Button bz) //判断并纪录每次点击按钮的信息
{ //当两次的按钮相同才能消去
if((s.trim()).equals("no"))
{
s=b1.getLabel();
x0=w1;
y0=w2;
n1=d[x0][y0];
b=bz;
x=w1;
y=w2;
n2=d[x][y];
ba=bz;
}
else
{
x0=x;
y0=y;
n1=d[x0][y0];
b=ba;
x=w1;
y=w2;
n2=d[x][y];
ba=bz;
if(n1==n2 ba!=b)
{
xiao();
}
}
}
public void xiao() //这里是整个游戏最重要的部分，就是判断两个按钮在信息
{ //相同的情况下能不能消去。仔细分析，不一条条注释
int i=0, j=0,n=0,k=0;
if((x0==x (y0==y+1||y0==y-1)) || ((x0==x+1||x0==x-1)(y0==y))) //相邻的情况
{
ba.setVisible(false);
b.setVisible(false);
guli();
s="no";
d[x0][y0]=0;
d[x][y]=0;
}
else
{
for (j=0;j7;j++ ) //两个按钮按行分析，看能否消去
{
if (d[x0][j]==0)
{
if (yj)
{
for (i=y-1;i=j;i-- )
{
if (d[x][i]!=0)
{
k=0;
break;
}
else
{
k=1;
}
}
if (k==1)
{
if (y0j)
{
for (i=y0-1;i=j ;i-- )
{
if (d[x0][i]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
if (y0j)
{
for (i=y0+1;i=j ;i++)
{
if (d[x0][i]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
}
}
if (yj)
{
for (i=y+1;i=j ;i++ )
{
if (d[x][i]!=0)
{
k=0;
break;
}
else
{
k=1;
}
}
if (k==1)
{
if (y0j)
{
for (i=y0-1;i=j ;i-- )
{
if (d[x0][i]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
if (y0j)
{
for (i=y0+1;i=j ;i++)
{
if (d[x0][i]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
}
}
if (y==j )
{
if (y0j)
{
for (i=y0-1;i=j ;i-- )
{
if (d[x0][i]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
if (y0j)
{
for (i=y0+1;i=j ;i++)
{
if (d[x0][i]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
}
}
if (k==2)
{ if (x0==x)
{
b.setVisible(false);
ba.setVisible(false);
guli();
s="no";
k=0;
d[x0][y0]=0;
d[x][y]=0;
}
if (x0x)
{
for (n=x0;n=x-1;n++ )
{
if (d[n][j]!=0)
{
k=0;
break;
}
if(d[n][j]==0 n==x-1)
{
b.setVisible(false);
ba.setVisible(false);
guli();
s="no";
k=0;
d[x0][y0]=0;
d[x][y]=0;
}
}
}
if (x0x)
{
for (n=x0;n=x+1 ;n-- )
{
if (d[n][j]!=0)
{
k=0;
break;
}
if(d[n][j]==0 n==x+1)
{
b.setVisible(false);
ba.setVisible(false);
guli();
s="no";
k=0;
d[x0][y0]=0;
d[x][y]=0;
}
}
}
}
}
for (i=0;i8;i++ ) //按列分析，看能不能消去
{
if (d[i][y0]==0)
{
if (xi)
{
for (j=x-1;j=i ;j-- )
{
if (d[j][y]!=0)
{
k=0;
break;
}
else
{
k=1;
}
}
if (k==1)
{
if (x0i)
{
for (j=x0-1;j=i ;j-- )
{
if (d[j][y0]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
if (x0i)
{
for (j=x0+1;j=i;j++ )
{
if (d[j][y0]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
}
}
if (xi)
{
for (j=x+1;j=i;j++ )
{
if (d[j][y]!=0)
{
k=0;
break;
}
else
{
k=1;
}
}
if (k==1)
{
if (x0i)
{
for (j=x0-1;j=i ;j-- )
{
if (d[j][y0]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
if (x0i)
{
for (j=x0+1;j=i ;j++ )
{
if (d[j][y0]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
}
}
if (x==i)
{
if (x0i)
{
for (j=x0-1;j=i ;j-- )
{
if (d[j][y0]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
if (x0i)
{
for (j=x0+1;j=i ;j++ )
{
if (d[j][y0]!=0)
{
k=0;
break;
}
else
{
k=2;
}
}
}
}
}
if (k==2)
{
if (y0==y)
{
b.setVisible(false);
ba.setVisible(false);
guli();
s="no";
k=0;
d[x0][y0]=0;
d[x][y]=0;
}
if (y0y)
{
for (n=y0;n=y-1 ;n++ )
{
if (d[i][n]!=0)
{
k=0;
break;
}
if(d[i][n]==0 n==y-1)
{
b.setVisible(false);
ba.setVisible(false);
guli();
s="no";
k=0;
d[x0][y0]=0;
d[x][y]=0;
}
}
}
if (y0y)
{
for (n=y0;n=y+1 ;n--)
{
if (d[i][n]!=0)
{
k=0;
break;
}
if(d[i][n]==0 n==y+1)
{
b.setVisible(false);
ba.setVisible(false);
guli();
s="no";
k=0;
d[x0][y0]=0;
d[x][y]=0;
}
}
}
}
}
}
}
}
求任何可在VC/C++环境下运行的连连看游戏完整源代码文件 谢谢了。。
#include stdio.h
#include conio.h
#include windows.h
#include time.h
#define Height 25 //迷宫的高度，必须为奇数
#define Width 25 //迷宫的宽度，必须为奇数
#define Wall 1
#define Road 0
#define Start 2
#define End 3
#define Esc 5
#define Up 1
#define Down 2
#define Left 3
#define Right 4
int map[Height+2][Width+2];
void gotoxy(int x,int y) //移动坐标
{
COORD coord;
coord.X=x;
coord.Y=y;
SetConsoleCursorPosition( GetStdHandle( STD_OUTPUT_HANDLE ), coord );
}
void hidden()//隐藏光标
{
HANDLE hOut = GetStdHandle(STD_OUTPUT_HANDLE);
CONSOLE_CURSOR_INFO cci;
GetConsoleCursorInfo(hOut,cci);
cci.bVisible=0;//赋1为显示，赋0为隐藏
SetConsoleCursorInfo(hOut,cci);
}
void create(int x,int y) //随机生成迷
{
int c[4][2]={0,1,1,0,0,-1,-1,0}; //四个方向
int i,j,t;
//将方向打乱
for(i=0;i4;i++)
{
j=rand()%4;
t=c[i][0];c[i][0]=c[j][0];c[j][0]=t;
t=c[i][1];c[i][1]=c[j][1];c[j][1]=t;
}
map[x][y]=Road;
for(i=0;i4;i++)
if(map[x+2*c[i][0]][y+2*c[i][1]]==Wall)
{
map[x+c[i][0]][y+c[i][1]]=Road;
create(x+2*c[i][0],y+2*c[i][1]);
}
}
int get_key() //接收按键
{
char c;
while(c=getch())
{
if(c==27) return Esc; //Esc
if(c!=-32)continue;
c=getch();
if(c==72) return Up; //上
if(c==80) return Down; //下
if(c==75) return Left; //左
if(c==77) return Right; //右
}
return 0;
}
void paint(int x,int y) //画迷宫
{
gotoxy(2*y-2,x-1);
switch(map[x][y])
{
case Start:
printf("入");break; //画入口
case End:
printf("出");break; //画出口
case Wall:
printf("▇");break; //画墙
case Road:
printf(" ");break; //画路
}
}
void game()
{
int x=2,y=1; //玩家当前位置，刚开始在入口处
int c; //用来接收按键
while(1)
{
gotoxy(2*y-2,x-1);
printf("●"); //画出玩家当前位置
if(map[x][y]==End) //判断是否到达出口
{
gotoxy(30,24);
printf("到达终点，按任意键结束");
getch();
break;
}
c=get_key();
if(c==Esc)
{
gotoxy(0,24);
break;
}
switch(c)
{
case Up: //向上走
if(map[x-1][y]!=Wall)
{
paint(x,y);
x--;
}
break;
case Down: //向下走
if(map[x+1][y]!=Wall)
{
paint(x,y);
x++;
}
break;
case Left: //向左走
if(map[x][y-1]!=Wall)
{
paint(x,y);
y--;
}
break;
case Right: //向右走
if(map[x][y+1]!=Wall)
{
paint(x,y);
y++;
}
break;
}
}
}
int main()
{
int i,j;
srand((unsigned)time(NULL)); //初始化随即种子
hidden(); //隐藏光标
for(i=0;i=Height+1;i++)
for(j=0;j=Width+1;j++)
if(i==0||i==Height+1||j==0||j==Width+1) //初始化迷宫
map[i][j]=Road;
else map[i][j]=Wall;
create(2*(rand()%(Height/2)+1),2*(rand()%(Width/2)+1)); //从随机一个点开始生成迷宫，该点行列都为偶数
for(i=0;i=Height+1;i++) //边界处理
{
map[i][0]=Wall;
map[i][Width+1]=Wall;
}
for(j=0;j=Width+1;j++) //边界处理
{
map[0][j]=Wall;
map[Height+1][j]=Wall;
}
map[2][1]=Start; //给定入口
map[Height-1][Width]=End; //给定出口
for(i=1;i=Height;i++)
for(j=1;j=Width;j++) //画出迷宫
paint(i,j);
game(); //开始游戏
getch();
return 0;
}
求c语言的连连看源程序
参考：
;vw=allssid=from=bd_page_type=1uid=D707A757C67E11B4F7ACD4D0D2C9C212pu=rc@1,pic@on,sl@1,pw@1000,sz@224_220,pd@1,fz@2,lp@0,tpl@color,st=1wk=rdmaxpage=6pos=all
以下是部分代码：
/*
* 连连看游戏C语言源代码
*/
#include stdio.h
#include graphics.h
#include stdlib.h
#include math.h
#include dos.h
#define true 1
#define false 0
/* ---------------------全局变量------------------------------------ */
int BkGndColor=BLACK;
int BorderColor=LIGHTGRAY;
int LineColor=LIGHTBLUE;/* 消除一对方块时时候的连线颜色 */
/* Pb - ProgressBar */
int PbColor=LIGHTGREEN;
int PbY=4;
int PbHeight=4;
int PbValue; /* 进度条百分比，初始值为100.*/
long StartTime; /* 开始时间的秒数，只统计分钟，秒 */
long TotalTime; /* 游戏总共的最大秒数！，*/
/* BoardDatas: a small-size board */
/* Board[x][y][0] - 0:empty, 1:filled */
/* Board[x][y][1] - cell's key; */
unsigned char Board[10][10][2];
int CellSize=30;
int BoardX=20;
int BoardY=60;
int BoardWidth=10;
int BoardHeight=10;
int CellColor=WHITE;
int SelColor=BLUE; /* selCell's border rect color */
int CurColor=RED; /* curCell's border rect color */
int EraColor=CYAN; /* 用于擦除cell的颜色！*/
int PairsCount; /* how much pairs we have put on board */
/* 用于存储逻辑坐标（索引） */
typedef struct _tagCELL
{
char x;
char y;
} CELL;
CELL selCell,curCell;/*缓存前一个被选中的位置以及当前所处位置！*/
/*Scan Codes Define*/
enum KEYCODES
{
K_ESC =0x011b,
K_UP =0x4800, /* upward arrow */
K_LEFT =0x4b00,
K_DOWN =0x5000,
K_RIGHT =0x4d00,
K_SPACE =0x3920,
K_P =0x1970,
K_RETURN =0x1c0d, /* Enter */
};
/* ---------------------函数列表------------------------------------ */
void InitGame(char *bgiPath);
void PlayGame();
void QuitGame();
void InitProgressBar();
void UpdateProgressBar(int percent);
void DrawCell(int key,int x,int y,int color);
void EraseCell(int x,int y);
void DrawBorderRect(CELL *c,int color);
void DrawGameOver(char* info);
int GetKeyCode();
int FindPath(CELL *c1,CELL *c2);
/*绘制消除方块时候的连接路径！，用指定颜色！*/
void DrawPath(int x1,int y1,int x2,int y2,int x3,int y3,int x4,int y4,int color);
/* ----------------------函数实现----------------------------------- */
/* ----------------------[ 核心算法 ]---------------------------------
* 先进行水平方向判断，找出两点所在的水平直线活动范围，
* 算出这两条线段在垂直方向的共同区域!!!，
* 遍历该区域判断能否在两线段间架起公垂线，能则两点连接上；
* 接着进行垂直方向判断，类同。无论两点在不在一条直线上，
* 都能使用该算法，因为两点同线只是两点作为矩形对角点的特例而已。
*/
/* 找到两个CELL之间的路径，成功返回true */
int FindPath(CELL *c1,CELL *c2)
{
int i,j,
path,min1,max1,min2,max2,left,right,top,bottom;
/*---------------（0）判断是否点中相同块！ ------------*/
if(Board[c1-x][c1-y][1] != Board[c2-x][c2-y][1])
return false;
/*---------------（1）查找水平方向公共区域！-----------*/
min1=max1=c1-x;
min2=max2=c2-x;
while(min1-1=0 Board[min1-1][c1-y][0]==0) min1--;
while(min2-1=0 Board[min2-1][c2-y][0]==0) min2--;
left=max(min1,min2); /* 左边界 */
while(max1+1BoardWidth Board[max1+1][c1-y][0]==0) max1++;
while(max2+1BoardWidth Board[max2+1][c2-y][0]==0) max2++;
right=min(max1,max2); /* 右边界 */
/* 检查两条水平线之间是否有公垂线连通！*/
/* 可以在边缘连通 */
if(left==0)
{
/* 左边缘连通 */
DrawPath(c1-x,c1-y, -1,c1-y, -1,c2-y, c2-x,c2-y, LineColor);
delay(6000);
DrawPath(c1-x,c1-y, -1,c1-y, -1,c2-y, c2-x,c2-y, BkGndColor);/*插除线条！*/
return true;
}
if(right==(BoardWidth-1))
{
DrawPath(c1-x,c1-y, BoardWidth,c1-y, BoardWidth,c2-y, c2-x,c2-y, LineColor);
delay(6000);
DrawPath(c1-x,c1-y, BoardWidth,c1-y, BoardWidth,c2-y, c2-x,c2-y, BkGndColor);/*插除线条！*/
return true;
}
for(i=left;i=right;i++)
{
path=0;/*统计垂直的公垂线长度！*/
for(j=min(c1-y,c2-y)+1;jmax(c1-y,c2-y);j++)
{
path+=Board[i][j][0];
if(path0) break;
}
if(path==0)
{
DrawPath(c1-x,c1-y, i,c1-y, i,c2-y, c2-x,c2-y, LineColor);
delay(6000);
DrawPath(c1-x,c1-y, i,c1-y, i,c2-y, c2-x,c2-y, BkGndColor);/*插除线条！*/
return true;
}
}
/*---------------（2）查找垂直方向公共区域！-----------*/
min1=max1=c1-y;
min2=max2=c2-y;
while(min1-1=0 Board[c1-x][min1-1][0]==0) min1--;
while(min2-1=0 Board[c2-x][min2-1][0]==0) min2--;
top=max(min1,min2);
while(max1+1BoardHeight Board[c1-x][max1+1][0]==0) max1++;
while(max2+1BoardHeight Board[c2-x][max2+1][0]==0) max2++;
bottom=min(max1,max2);
/* 检查两条垂直线之间是否有公垂线连通！*/
/* 可以在边缘连通 */
if(top==0)
{
/* 同在顶端消除 */
DrawPath(c1-x,c1-y, c1-x,-1, c2-x,-1, c2-x,c2-y, LineColor);
delay(6000);
DrawPath(c1-x,c1-y, c1-x,-1, c2-x,-1, c2-x,c2-y, BkGndColor);/*插除线条！*/
return true;
}
if(bottom==(BoardHeight-1))
{
DrawPath(c1-x,c1-y, c1-x,BoardHeight, c2-x,BoardHeight, c2-x,c2-y, LineColor);
delay(6000);
DrawPath(c1-x,c1-y, c1-x,BoardHeight, c2-x,BoardHeight, c2-x,c2-y, BkGndColor);/*插除线条！*/
return true;
}
谁能帮忙写一个c语言连连看游戏，要4x4的，功能最简单的就可以！但是要实现连连看应该有的功能
刚写的，新鲜出炉：
#include stdio.h
#include string.h
#include stdlib.h
#include time.h
#define MAPSIZE 4
#define MAXLINESIZE 60
//typedef enum{false,true} bool;
typedef struct{
int x,y;
}Point;
const char pictureTable[]={" ABCEDFGHI"};
bool judgeLine(char **MP,Point *start,Point *end){
int i;
if(start-x==end-x){
if(start-y end-y){
for(i = start-y - 1 ; i end-y ; i--)
if(MP[start-x][i]!=0)
return false;
return true;
}
else{
for(i = start-y + 1 ; i end-y ; i++)
if(MP[start-x][i]!=0)
return false;
return true;
}
}
else if(start-y==end-y){
if(start-x end-x ){
for(i = start-x - 1 ; i end-x ; i--)
if(MP[i][start-y]!=0)
return false;
return true;
}
else{
for(i = start-x + 1 ; i end-x ; i++)
if(MP[i][start-y]!=0)
return false;
return true;
}
}
return false;
}
bool judgeTwoLines(char **MP,Point *start,Point *end,Point *mid){
Point p1,p2;
mid-x=-1;
mid-y=-1;
if(judgeLine(MP,start,end)==true) return true;
p1.x=start-x;
p1.y=end-y;
p2.x=end-x;
p2.y=start-y;
mid-x=p1.x;
mid-y=p1.y;
if(MP[p1.x][p1.y]==0 judgeLine(MP,start,p1) judgeLine(MP,end,p1)) return true;
mid-x=p2.x;
mid-y=p2.y;
if(MP[p2.x][p2.y]==0 judgeLine(MP,start,p2) judgeLine(MP,end,p2)) return true;
return false;
}
bool judgeTreeLines(char **MP,Point *start,Point *end,Point *mid1,Point *mid2,int n){
int i;
mid1-x=-1;mid1-y=-1;
mid2-x=-1;mid2-y=-1;
if(judgeTwoLines(MP,start,end,mid1)) return true;
for( i=start-x - 1;i=0;i--){
if(MP[i][start-y]!=0) break;
mid1-x=i;
mid1-y=start-y;
if(judgeTwoLines(MP,mid1,end,mid2)) return true;
}
for( i=start-x + 1;i=n+1;i++){
if(MP[i][start-y]!=0) break;
mid1-x=i;
mid1-y=start-y;
if(judgeTwoLines(MP,mid1,end,mid2)) return true;
}
for( i=start-y - 1;i=0;i--){
if(MP[start-x][i]!=0) break;
mid1-x=start-x;
mid1-y=i;
if(judgeTwoLines(MP,mid1,end,mid2)) return true;
}
for( i=start-y + 1;i=n+1;i++){
if(MP[start-x][i]!=0) break;
mid1-x=start-x;
mid1-y=i;
if(judgeTwoLines(MP,mid1,end,mid2)) return true;
}
return false;
}
void ptMap(char **MP,int n){
int space=(MAXLINESIZE-n*2)/2;
int i,j;
for(i=0;i(MAXLINESIZE-10)/2;i++)
printf(" ");
printf("《连连看》\n");
for(i=2;ispace;i++) printf(" ");
printf("x\n");
for(i=1;i=n;i++){
for(j=2;jspace;j++)
printf(" ");
printf("%d ",i);
for(j=1;j=n;j++)
printf("%c ",pictureTable[MP[i][j]]);
printf("\n");
}
for(i=0;ispace;i++)
printf("*");
for(i=0;in;i++)
printf("%d*",i+1);
for(i=1;ispace;i++)
printf("*");
printf("\n");
}
char **createMap(int n){
char **ret;
int i;
ret=(char**)malloc(sizeof(char*)*(n+2));
for(i=0;in+2;i++)
ret[i]=(char*)malloc(sizeof(char)*(n+2));
return ret;
}
void ranMap(char **MP,int n){
int *all=(int*)malloc(sizeof(int)*n*n);
int i,tmpi,tmp;
for(i=0;in*n;i++)
all[i]=i/4+1;
for(i=0;in*n;i++){
tmpi=rand()%(n*n-i);
tmp=all[tmpi];
all[tmpi]=all[n*n-i-1];
all[n*n-i-1]=tmp;
}
for(i=0;in+2;i++){
MP[0][i]=0;
MP[n+1][i]=0;
MP[i][0]=0;
MP[i][n+1]=0;
}
tmpi=0;
for(i=1;i=n;i++)
for(tmp=1;tmp=n;tmp++)
MP[i][tmp]=all[tmpi++];
}
void deletePoints(char **MP,Point *p1,Point *p2){
MP[p1-x][p1-y]=0;
MP[p2-x][p2-y]=0;
}
int playTurns(int n){
int rest=n*n;
char **mp=createMap(n),c;
ranMap(mp,n);
Point mid1,mid2,pt1,pt2;
while(1){
ptMap(mp,n);
printf("请输入消去的坐标1(x1 y1):\n");
scanf("%d%d",pt1.x,pt1.y);
printf("请输入消去的坐标2(x2 y2):\n");
scanf("%d%d",pt2.x,pt2.y);
if((pt1.x==pt2.x pt1.y==pt2.y) || (pt1.x1 || pt1.xn || pt2.x 1 || pt2.x n || pt1.y1 || pt1.yn || pt2.y 1 || pt2.y n)){
printf("无法消除这两图案，请再次检查。");
}
else if(mp[pt1.x][pt1.y]!=0 mp[pt1.x][pt1.y]==mp[pt2.x][pt2.y] judgeTreeLines(mp,pt1,pt2,mid1,mid2,n)){
if(mid1.x==-1){
printf("Direct\n");
}
else if(mid2.x==-1){
printf("TwoLines :(%d,%d)\n",mid1.x,mid1.y);
}
else{
printf("TreeLines:(%d,%d)(%d,%d)\n",mid1.x,mid1.y,mid2.x,mid2.y);
}
deletePoints(mp,pt1,pt2);
printf("消去成功！\n");
rest-=2;
if(rest==0){
printf("恭喜！你已消去所有图案！\n");
break;
}
}
else{
printf("无法消除这两图案，请再次检查。");
}
printf("继续游戏(N/n不继续)？");
scanf(" %c",c);
if(c=='N' || c=='n') break;
}
printf("是否重新开局(Y/y继续)？");
scanf(" %c",c);
if(c=='y' || c=='Y') return 1;
return 0;
}
int main(){
srand(time(0));
while(playTurns(4));
return 0;
}
急需一个用C语言编写的 连连看游戏 代码啊，需要简易一点的，适合刚入门新手看懂的
你这10财富也太少了，这种范式代码很多好吧 自己百度下 一搜一大堆
guli(); s="no"; k=0; d[x0][y0]=0; d[x][y]=0; } if (y0y) { for (n=y0;n=y-1 ;n++ ) { if
) { wei(4,2,b17); } }); b18.addMouseListener(new MouseAdapter(){ public void mouseClicke
c1-y][1] != Board[c2-x][c2-y][1])return false;/*---------------（1）查找水平方向公共区域！-------